Connect with us

Driver for IGBT? (-ve off and +ve on from CMOS)

Discussion in 'Electronic Basics' started by Richard, Oct 10, 2005.

  1. Richard

    Richard Guest

    I would like to drive an IGBT from a microcontroller (CMOS).
    Is there a driver circuit that can outputs around +15v (on) and a -5v (off)
    for switching frequencies around 40Khz.
    I think -ve voltage supply is avaliable
    What is this type of circuit called?

    Thanks
    Richard
     
  2. I am also interested for something that makes negative off
    signal... If you find anything, let me know.

    i
     
  3. Mook Johnson

    Mook Johnson Guest

    I'm not sure why you'd need to drive the gate negative. below the threshold
    voltage should be fine to turn it off (just like a Mosfet). if you insist
    on turning swingint it negative, you might look at a gate drive transformer
    that naturally swings negative (ac coupling the gate signal)

    A discrete design wouldn't be to difficult to come up with.
     
  4. Terry Given

    Terry Given Guest

    because negative gate bias can prevent dV/dt thru Cmiller from turning
    the darned thing back on again.

    IR wrote a nice paper about 6-7 years ago, where they go through and
    show at what point -ve bias becomes necessary. its mandatory for
    man-sized IGBTs, but baby IGBTs can often do without.
     
  5. I found this some where, I forget where, but its a novel approach.
    A MAX232, +/-10 volt swing.

    Cheers
     
  6. Robert Baer

    Robert Baer Guest

    I would never even *think* of Maxim or Dallas VaporWare.
     
  7. Bob Monsen

    Bob Monsen Guest

    Those MAX232 chips have been around for a while, so they probably aren't
    too dangerous to depend on (although I've had my run-ins with Maxim and
    Dallas). However, getting them to drive an IGBT at 40kHz may not be
    possible, due to the large input capacitance, which may require lots of
    current to overcome quickly. They use a charge pump, which may not be able
    to supply the needed current.

    There are dedicated IGBT driver chips available. There are also integrated
    H-Bridge chips that would work nicely. Using one of those is probably the
    simplest and safest course.
     
  8. Fred Bloggs

    Fred Bloggs Guest

    I would like to drive an IGBT from a microcontroller (CMOS).
    Surprisingly it is called a gate driver circuit. IGBTs imply large power
    switching so that even though the switching frequency is a modest 40KHz,
    the driver is still required to switch the device through ON/OFF
    transitions quickly to reduce losses and component heating. Micrel at
    http://www.micrel.com manufactures a line of powerful gate driver
    products capable of handling nearly any situation. The most basic part
    in their line-up is the MIC4120, a non-inverting driver capable of
    supporting 6A peak gate drive currents. A high speed interface to a
    microcontroller might look like this:
    View in a fixed-width font such as Courier.
     
  9. Paul Burke

    Paul Burke Guest

    Remember the early ones, that typically lasted about 3 months? They
    later changed the circuit, so that the high side output capacitor went
    to 5V instead of ground. Before that, it would latch up unpredicatbly,
    usually when you'd just delivered a lot to an important customer.

    Paul Burke
     
  10. Check out Fuji's gate driver:

    http://www.fujielectric.co.jp/eng/fdt/scd/pdf/Manual/REH924.pdf

    Apparently, it makes negative reverse bias of -5V and has a built in
    opto coupler.

    i
     
  11. Ken Smith

    Ken Smith Guest

    Use an LT1081. They are easy to get. I think they are about as slow and
    weak as the MAX232 so it really isn't such a good idea.
     
  12. On Mon, 10 Oct 2005 14:05:58 +0000 (UTC), the renowned
    MAX232s are multiple-sourced these days, so you're not stuck with
    Maxim, any more than you need fear the uA709A because you don't trust
    Fairchild. ;-)


    Best regards,
    Spehro Pefhany
     
  13. For big IGBTs, some negative gate drive is an advantage.
    There's a lot of capacitance.


    Best regards,
    Spehro Pefhany
     
  14. I am actually playing with IGBTs.

    I have Toshiba-MG200Q2YS40 IGBT.

    It does NOT turn off itself if I simply remove Gate/Emitter voltage.

    I have to apply reverse voltage to turn it off. I actually tried it.

    It would, possibly, switch itself off with a small resistor between
    gate and emitter, but it is just a hypothesis of mine.


    i
     
  15. Jamie

    Jamie Guest

    i really don't understand why you need that?, normally pulling the
    gate low will do how ever if you insist.
    simply provide a +15 volt source via a resistor sufficient enough
    to drive the gate in the on mode, use something like a simply NPN
    transistor to pull it low at the gate to shut the IGBT off, put the
    emitter to a - supply.
    since these Gates are H Z, the - source does not need to be much.
    using either a high freq OSC from something simple like a 555 or
    an output from the MC, you can generate - voltage via a Cap and diode et..
    the base of the transistor will get switched from the MC signal line.
     
  16. Fred Bloggs

    Fred Bloggs Guest

    Gate charge is what you really mean...capacitance can be misleading
    because it ignores threshold voltage. After looking at that Fugi
    Electric line, it seems they require hundreds of ohms of gate series
    damping resistance- because of the bipolar current gain multiplying the
    lead inductance reflected into the gate circuit most likely- and this
    explains the negative bias requirement, as a lower impedance pulldown
    would shunt the Miller transient- at turn-off.
     
  17. Terry Given

    Terry Given Guest

    a decent short from gate to emitter would do it.
    what you are loking at is a combination of things - miller effect
    mostly. There is a non-linear capacitance between gate and collector,
    call it Ccg. Whenever the collector voltage changes value (say from
    Vcesat to Vdc when turning off) the voltage slews at some rate, dV/dt.
    This causes current to flow thru Ccg, Icg = Ccg*dVce/dt. This current
    flows into the gate circuit, charging up Cge. If the gate driver is not
    very low impedance, Icg will happily charge Cge up to Vth, and beyond.
    when turning the IGBT on, Vce falls so Icg flows out of Cge, discharging
    it - IOW trying to turn it back off.

    This can greatly increase switching time (and hence losses) and, if the
    gatedrive is bad enough, can make the device "latch" in the on-state,
    with Vge sitting around Vth - but you need a seriously piss-weak gate
    driver to do that. It can even over-voltage the gate, poke a hole thru
    the thin gate oxide layer, and KABOOM, one dead IGBT.

    a negative gate bias means the switch-off miller current has to supply
    more charge to reach Vth - instead of supplying Qg = Vth*Cge, it must
    supply (Vth + |Vnegative|)*Cge. This allows the use of higher gate
    impedance circuitry. Its not just the resistance, its the inductance
    too. bad gate drives have high resistance and high inductance; good gate
    drives have low resistance and low inductance.

    to give you an idea, I've worked on gatedrives for 0.5kW - 2.2kW drives
    that have about -3V reverse-bias, and -15V for 400kW drives (6 300A fuji
    IGBTs directly paralleled).

    modern IGBT designs have much better packaging - far lower inductance,
    both Lce and Lge. with some of these IGBTs it is theoretically possible
    to keep the gate impedance sufficiently low that negative bias isnt
    necessary. Like I said, IR wrote a pretty good paper on the topic - why
    you dont, in theory, need -ve bias for little IGBTs.

    In practice, if you screw up the gatedriver, your circuit is going to go
    BANG. If you really know what you are doing, and are very good at taking
    measurements, its fairly easy to dewsign gatedrivers and make them work.
    OTOH if you are not too experienced, its very easy to destroy many, many
    IGBTs.

    Generally the gatedrive is a lot cheaper than the IGBTs, so its sensible
    to throw -ve bias at it. For little IGBTs, it can be as simple as a
    zener in series with the gate resistor, with a 100nF cap across it.

    Cheers
    Terry
     
  18. Terry, it is great that you participate in this thread and actually
    have experience with power switching. I am building a DC -> AC
    inverter for a TIG welder and seems like I can learn a lot from you.

    This is a 200A constant current, 28 V welding current, 80 v open
    circuit voltage tig welder.I have aforementioned IGBTs.

    Can you give a little schematic of how go give the gates negative bias
    during turn off. I definitely want to turn IGBTs off in the best
    possible way. I want the simplest solution.

    Another question, would a resistor between gate and emitter solve ths
    issue? Or is that not the case?

    thanks a lot for your time!

    i
     
  19. Terry Given

    Terry Given Guest

    here's one simple solution: use an optocoupler to drive a FET driver
    chip, running from a +10V/-5V supply. connect the supply 0V to the IGBT
    emitter, +10V to the driver chip Vcc and -5V to the driver chip gnd.
    that way the driver chip "sees" a 15V supply, whereas the IGBT "sees"
    +10V and -5V.

    we used to use tens of thousands of UC3842 smps chips as gate drivers.
    An opto drove the 3842, which ran from a 15V isolated supply. the 3842
    Vref output was connected to the IGBT emitter, giving +10V/-5V gate
    drive. We had hefty caps from +10V to 0V to -5V. for little drives. for
    big drives, we used +/-15V supplies.

    for really little drives (< 2kW) we used the series zener trick. get a
    15V FET driver, with a 15V supply. bung a 5V zener in series with the
    output, with a 100nF cap across it. When the gatedrive output is high,
    5V is dropped across the zener, charging the cap - the cathode end of
    the cap is 5V below the anode end, and the gate gets 10V.

    when the gatedrive output switches to 0V, the anode end of the cap is at
    0V. The cathode end of the cap is 5V more negative than the anode end,
    so the gate sees -5V. The 100nF cap needs to be at least 10x the gate
    capacitance. For larger IGBTs, 1uF would be better.

    this is probably the easiest option for you, and works with *any* gate
    driver. Make sure you use a stompy zener though - you dont want its
    series impedance to be large compared to your gate resistor. Think BZT03
    or suchlike.

    gatedrives are tricky, and until they work you kill a lot of IGBTs. why
    not just buy one from Semikron.
    yes but no. You would need a seriously low valued resistor (< 10R) to
    even attempt to hold the IGBT off wrt Cmiller, which would quite
    successfully bugger up the gatedrive. not to mention the 15V^2/10R = 22W
    or more power dissipation when the gatedrive is ON.
    Cheers
    Terry
     
  20. That's good to know. I am looking at using an IR21094 chip. Is it true
    that it supports that configuration (it seems to be the case to me,
    but I want to confirm).
    I will double check, but I think that I have a 12V with +-5V power
    supply lying around. If not, I can make one from something, even a
    dual power supply for starters.

    Now, I want to make sure that I understand what you mean.

    Question: So, the above should work for a "big drive" (which, I
    assume, applies to my 200 A system). What you write below is simply
    another option that applies to "smaller drives".

    Is that correct?
    I think that I understand.
    Which one would you suggest?
    Not very feasible... I was hoping to be able to get away with a much
    smaller resistor...

    Thanks a lot...

    i

    --
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-